Existence and construction of isomorphism between finite groups

Question

Assume I have two finite groups $G$ and $H$ of equal order. Further assume I have found minimal generating sets $A$ and $B$ for a the two groups respectively (of equal size) and additionally (see comments) I know at least one decomposition into generating elements of all $g \in G$.

I now want to find out if the two groups are isomorphic give this extra information of the minimal generating sets.

Is there an approach along these lines:

Define a bijective function $f: A\to B$. Extend the function to all of $G$ in the following manner:
For $g \in G$ find a decomposition of $g = a_1 + \cdots + a_m$ where $a_i \in A$ and define $f(g) = \sum_{i=1}^m f(a_i)$. If this extension fulfills the homomorphism property $f(g_1 + g_2) = f(g_1) + f(g_2)$ for all $g_1, g_2 \in G$ then it is an isomorphism.

Question 1. Is the above statement correct? Do I need to check $f(G) = H$ or is this already implicitly true?

Question 2. Here I need to expand the function for all $g \in G$ and check the homomorphism property for all pairs of elements from $G$.
Given the information of two minimal generating sets, can I reduce the amount of checking I have to do?
(Checking only the pairs of generators is obviously not enough since the extension fulfills the homomorphism property for elements from $A$ by construction)

Answer 1

(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)

There are two things you seem to be missing.

1. If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
2. The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $\mathbb{Z}_3\times\mathbb{Z}_{18}$. This is isomorphic to $\mathbb{Z}_6\times\mathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.

The following is then true:

Theorem. Let $A$ be a minimal generating set for $G$, and let $\{B_i\}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $G\cong H$ if and only if there exists a bijection $f_{i, j}: A\rightarrow B_i$ which extends to an isomorphism $G\rightarrow H$ in the way you describe.

Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.

*If $G$ is given by a presentation $\langle \mathbf{x}\mid\mathbf{r}\rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $R\in \mathbf{r}$.